chapter 12 frequency response analysis (bode ... shoubra...• plot asymptotic approximations to the...

Electrical Engineering Department Dr. Ahmed Mustafa Hussein

Benha University Faculty of Engineering at Shubra

1 Chapter Twelve: Bode Plots Dr. Ahmed Mustafa Hussein

CHAPTER 12 FREQUENCY RESPONSE ANALYSIS (Bode Plots)

After completing this chapter, the students will be able to:

• Plot asymptotic approximations to the frequency response of an open-loop

control system,

• Use the Bode plot to determine the stability of open-loop systems

• Find the bandwidth, peak magnitude, and peak frequency of a closed-loop

frequency response.

1. Introduction

Frequency response methods, developed by Nyquist (1930) and Bode (1945), are older

than the root locus method, which was discovered by Evans in 1948.

Obtaining the frequency response from the transfer function by substituting the value

of (ω) directly in the system transfer function is a tedious task. The frequency range

required in frequency response is often so wide that it is inconvenient to use a linear

scale for the frequency axis. Also, there is a more systematic way of locating the

important features of the magnitude and phase plots of the transfer function. For these

reasons, it has become standard practice to use a logarithmic scale for the frequency




axis and a linear scale in each of the separate plots of magnitude and phase. Such

semi-logarithmic plots of the transfer function—known as Bode plots—have become

the industry standard. Bode plots contain the same information as the non-logarithmic

plots, but they are much easier to construct.

The transfer function GH(s) can be expressed as:

𝐺𝐻(𝑠) = |𝐺𝐻|∠𝜑

Since Bode plots are based on logarithms, it is important that we keep the following

properties of logarithms in mind:

log 𝑋1𝑋2 = log 𝑋1 + log 𝑋2

log 𝑋1/𝑋2 = log 𝑋1 − log 𝑋2

log 𝑋12 = 2 log 𝑋1

log 1 = 0

2. The Decibel Scale

In communications systems, gain is measured in Bels. The bel is used to measure the

ratio of two levels of power or power gain G; that is,

𝐺 = log𝑃1

𝑃2 𝐵𝑒𝑙𝑠

Deci is a suffix express 10 times of the quantity.

𝐺 = 10 × log𝑃1

𝑃2 𝑑𝑒𝑐𝑖𝐵𝑒𝑙𝑠

deciBels or (dB) provides less magnitude. Decibels is 1/10 of bels.

Consider the electric network shown in Fig. 1.

Fig. 1, Simple electric circuit

If P1 is the input power, P2 is the output (load) power, R1 is the input resistance, and R2

is the load resistance, then:




𝑃1 =𝑉1

2

𝑅1= 𝐼1

2𝑅1

𝑃2 =𝑉2

2

𝑅2= 𝐼2

2𝑅2

Assuming that R1 = R2 a condition that is often assumed when comparing voltage

levels, then:

𝐺𝑑𝐵 = 10 × log𝑃1

𝑃2= 10 × log (

𝑉1

𝑉2)

2

= 20 × log (𝑉1

𝑉2)

By the same way, assuming that R1 = R2 a condition that is assumed for comparing

current levels, then:

𝐺𝑑𝐵 = 10 × log𝑃1

𝑃2= 10 × log (

𝐼1

𝐼2)

2

= 20 × log (𝐼1

𝐼2)

To conclude the above information:

• 10 log is used for power, while 20 log is used for voltage or current, because of

the square relationship.

• The dB value is a logarithmic measurement of the ratio of one variable to

another of the same type. Therefore, it applies in expressing the transfer

function.

In Bode plots, the magnitude is plotted in Decibels (dB) versus frequency. The dB

quantity can be obtained as:

𝐺𝐻𝑑𝐵 = 20 log 𝐺𝐻

Moreover, the phase angle (φ) is plotted versus frequency. Both magnitude and phase

plots are made on semi-log graph paper.

3. Asymptotic Bode Plots (Open-Loop Frequency Response)

The log-magnitude and phase frequency response curves as functions of log ω are

called Bode plots or Bode diagrams. Sketching Bode plots can be simplified because

they can be approximated as a sequence of straight lines. Straight-line approximations

simplify the evaluation of the magnitude and phase frequency response.

Consider the following transfer function that may be written in terms of factors that

have real and imaginary parts such as:




𝐺(𝑗𝜔) = 𝐾 (1 +

𝑗𝜔𝑧1

) (1 +𝑗𝜔𝑧2

) {1 + 𝑗2𝜉1𝜔

𝜔𝑛+ (

𝜔𝜔𝑛

)2

} …

(𝑗𝜔)±1 (1 +𝑗𝜔𝑝1

) (1 +𝑗𝜔𝑝2

) {1 + 𝑗2𝜉2𝜔

𝜔𝑛+ (

𝜔𝜔𝑛

)2

} …

this is called the Bode (Standard) form of the system transfer function that may

contain seven different factors:

• Bode gain K

• Pole at origin (𝑗𝜔)−1 or zero at origin (𝑗𝜔)+1

• Real pole (1 +𝑗𝜔

𝑝1)

−1 and/or real zero (1 +

𝑗𝜔

𝑧1)

• Quadratic pole {1 + 𝑗2𝜉2𝜔

𝜔𝑛+ (

𝜔

𝜔𝑛)

2}

−1

or quadratic zero {1 + 𝑗2𝜉1𝜔

𝜔𝑛+ (

𝜔

𝜔𝑛)

2}

In constructing a Bode plot, we plot each factor separately and then combine them

graphically. The factors can be considered one at a time and then combined additively

because of the logarithms involved. For this mathematical convenience of the

logarithm, Bode plots is considered as a powerful engineering tool.

In the following subsections, we will make straight-line plots of the factors listed

above. These straight-line plots known as asymptotic (approximate) Bode plots.

3.1 Bode Gain

For the gain K, there are two cases:

K is +ve and less than one: the magnitude 20 log K is negative and the phase is 0◦;

K is +ve and greater than one: the magnitude 20 log K is positive and the phase is 0◦;

K is -ve and less than one: the magnitude 20 log K is negative and the phase is -180◦;

K is -ve and greater than one: the magnitude 20 log K is +ve and the phase is -180◦;

Both of the magnitude and phase are constant with frequency. Thus the magnitude and

phase plots of the gain are shown in Fig.2.




Fig. 2, Magnitude and phase plots of Bode gain

3.2 Zero at origin

For the zero (jω) at the origin, the magnitude is 20 log10 ω and the phase is 90◦. These

are plotted in Fig. 3, where we notice that the magnitude is represented by a straight

line with slope of 20 dB/decade and intersect the 0dB line at ω=1 and extended to

intersect the vertical axis. But the phase is represented by straight line parallel to

horizontal axis with constant value at 90.

ونمده ω 1 =ديسبل لكل ديكاد ويمر بخط الصفر ديسيبل عند 20القيمة تمثل بخط مستقيم ميله

ωحور درجة وتمثل بخط مستقيم موازى لم 90أما الزاوية فقيمتها ثابتة عند

Fig. 3, Magnitude and phase plots of zero at origin

In general, for multiple zeros at origin (jω)N, where N is an integer, the magnitude plot

will have a slope of (20×N) dB/decade. But the phase is (90×N) degrees.

3.3 Pole at origin

For the pole (jω)-1 at the origin, the magnitude is -20 log10 ω and the phase is -90◦.

These are plotted in Fig. 4, where we notice that the magnitude is represented by a

straight line with slope of -20 dB/decade and intersect the 0dB line at ω=1 and

extended to intersect the vertical axis. But the phase is represented by straight line

parallel to horizontal axis with constant value at -90.

ونمده ω 1 =د ويمر بخط الصفر ديسيبل عند يديسبل لكل ديك -20القيمة تمثل بخط مستقيم ميله

ωبخط مستقيم موازى لمحور درجة وتمثل -90أما الزاوية فقيمتها ثابتة عند




In general, for multiple poles at origin (jω)-N, where N is an integer, the magnitude

plot will have a slope of - (20×N) dB/decade. But the phase is - (90×N) degrees.

Fig. 4, Magnitude and phase plots of pole at origin

3.4 Real Zero

The magnitude of a real zero (1 +𝑗𝜔

𝑧1) is obtained from 20 𝑙𝑜𝑔 |1 +

𝑗𝜔

𝑧1|, and the phase

is obtained from 𝑡𝑎𝑛−1 (𝜔

𝑧1). We notice that:

- For small values of ω, the magnitude is 20 𝑙𝑜𝑔 |1 +𝑗𝜔

𝑧1| ≅ 20 log 1 = 0

- For large values of ω, the magnitude is 20 𝑙𝑜𝑔 |1 +𝑗𝜔

𝑧1| ≅ 20 log |

𝜔

𝑧1|

From the above two points, we can approximate the magnitude of real zero by two

straight lines (at ω → 0 : a straight line with zero slope with zero magnitude) and (at ω

→ ∞ : a straight line with slope 20 dB/decade). At the frequency ω = z1 where the two

asymptotic lines meet is called the corner frequency. Thus, the approximate magnitude

plot is shown in Fig. 5. The actual plot for real zero is also shown in that figure. Notice

that the approximate plot is close to the actual plot except at the corner frequency,

where ω = z1 and the deviation is 20 𝑙𝑜𝑔|1 + 𝑗1| ≅ 20 log √2 = 3 𝑑𝐵.

ولانمده ω =1Zديسبل لكل ديكيد ويمر بخط الصفر ديسيبل عند +20القيمة تمثل بخط مستقيم ميله

ونصل بينهما بخط مستقيم 90( الزاوية = 110Z( الزاوية = صفر، ثم ديكيد بعد )10/1Zقبل )ديكيد الزاوية:

درجة لكل ديكيد 45ليكون ميل الخط

Fig. 5, Magnitude and phase plots of real zero




The phase angle of real zero that given as 𝑡𝑎𝑛−1 (𝜔

𝑧1) is represented as a straight-line

approximation, φ = 0 for ω ≤ z1/10, φ = 45◦for ω = z1, and φ = 90◦ for ω ≥ 10z1 as

shown in Fig. 4. The straight line has a slope of 45 per decade.

For example, consider the real zero (S+1), it will be (1+jω) in the Bode form. Then:

𝑚𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒 = 20 log(√1 + 𝜔2) , 𝑝ℎ𝑎𝑠𝑒 = 𝑡𝑎𝑛−1𝜔

1

The following table shows the actual and asymptotic values of the magnitude and

phase of that real zero.




3.5 Real Pole

The magnitude of a real pole (1 +𝑗𝜔

𝑝1)

−1 is obtained from −20 𝑙𝑜𝑔 |1 +

𝑗𝜔

𝑝1|, and the

phase is obtained from −𝑡𝑎𝑛−1 (𝜔

𝑝1). We notice that:

- For small values of ω, the magnitude is −20 𝑙𝑜𝑔 |1 +𝑗𝜔

𝑝1| ≅ 20 log 1 = 0

- For large values of ω, the magnitude is −20 𝑙𝑜𝑔 |1 +𝑗𝜔

𝑝1| ≅ −20 log |

𝜔

𝑝1|

From the above two points, we can approximate the magnitude of real pole by two

straight lines (at ω → 0 : a straight line is with zero slope and zero magnitude) and (at

ω → ∞ : the straight line is with slope -20 dB/decade). At the frequency ω = p1 where

the two asymptotic lines meet is called the corner frequency. Thus, the approximate

magnitude plot is shown in Fig. 6. The actual plot for real pole is also shown in that

figure. Notice that the approximate plot is close to the actual plot except at ω = p1, the

deviation is −20 𝑙𝑜𝑔|1 + 𝑗1| ≅ −20 log √2 = −3 𝑑𝐵.

ولانمده ω =p1ديسبل لكل ديكيد ويمر بخط الصفر ديسيبل عند -20القيمة تمثل بخط مستقيم ميله

The phase angle of real pole that given as −𝑡𝑎𝑛−1 (𝜔

𝑝1) is represented as a straight-line

approximation, φ = 0 for ω ≤ p1/10, φ = -45◦for ω = p1, and φ = -90◦ for ω ≥ 10p1 as

shown in Fig. 6. The straight line has a slope of -45 per decade.

ونصل بينهما بخط مستقيم 90-( الزاوية = 10p1( الزاوية = صفر، ثم ديكيد بعد )p1/10الزاوية: ديكيد قبل )

درجة لكل ديكيد -45ميل الخط ليكون




Fig. 6, Magnitude and phase plots of real pole

3.6 Quadratic Zero

The magnitude of a quadratic zero {1 + 𝑗2𝜉2𝜔

𝜔𝑛+ (

𝜔

𝜔𝑛)

2} is obtained as

20 log |1 + 𝑗2𝜉2

𝜔

𝜔𝑛+ (

𝜔

𝜔𝑛)

2

| = 0 𝑓𝑜𝑟 𝜔 → 0

20 log |1 + 𝑗2𝜉2𝜔

𝜔𝑛+ (

𝜔

𝜔𝑛)

2| = 20𝑙𝑜𝑔 |(

𝜔

𝜔𝑛)

2| = 40 𝑙𝑜𝑔 |(

𝜔

𝜔𝑛)| 𝑓𝑜𝑟 𝜔 → ∞

Thus, the amplitude plot consists of two straight asymptotic lines: one with zero slope

for ω < ωn and the other one with slope −40 dB/decade for ω > ωn, with ωn as the

corner frequency. Figure 7 shows the approximate and actual amplitude plots. Note

that the actual plot depends on the damping ratio ξ2 as well as the corner frequency ωn.

The significant peaking in the neighborhood of the corner frequency should be added

to the straight-line approximation if a high level of accuracy is desired. However, we

will use the straight-line approximation for the sake of simplicity.

The phase plot is a straight line with a slope of 90◦ per decade starting at ωn/10 and

ending at 10ωn, as shown in Fig. 7. We see again that the difference between the

actual plot and the straight-line plot is due to the damping factor.




Fig. 7, Magnitude and phase plots of quadratic zero

For the quadratic pole {1 + 𝑗2𝜉2𝜔

𝜔𝑛+ (

𝜔

𝜔𝑛)

2}

−1

the plots shown in Fig. 7 are inverted

because the magnitude plot has a slope of -40 dB/decade while the phase plot has a

slope of -90◦ per decade.

Since the asymptotes are quite easy to draw and are sufficiently close to the exact

curve, the use of such approximations in drawing Bode diagrams is convenient in

establishing the general nature of the frequency-response characteristics quickly with a

minimum amount of calculation and may be used for most preliminary design work.

4. Closed-Loop Stability Analysis Using Bode Plots

The gain crossover frequency g is defined as the frequency at which the total

magnitude equals 0 dB. Therefore, its value can be determined from the intersection of

the total magnitude line with the 0 dB line as shown in Fig. 8. On the other hand, the

phase crossover frequency p is defined as the frequency at which the total phase

equals −180. Therefore, its value can be determined from the intersection of the total

phase line with the −180 line as shown in Fig. 8.




Fig. 8, Gain and Phase crossover frequencies

The system Gain Margin (GM) in dB is the additional gain that makes the system on

the edge of instability. GM can be determined by calculating the total magnitude at

= p. Also, the system Phase Margin PM in degrees is the additional phase that makes

the system on the edge of instability. PM can be determined by calculating the total

phase at = g as shown in Fig. 9.

Fig. 9, Gain and Phase Margins




5. Plotting Bode Plots Using Matlab

To specify the frequency range for Bode plots, use the command:

>> logspace(d1,d2)

This generate 50 points logarithmically equally spaced between 10d1 and 10d2.

For example, if we need Bode plot starts at 0.1 rad/sec and finish at 100 rad/sec, enter

the command:

>> logspace(-1,2)

If we need to change the number of points between d1 and d2 rather than 50, use the

command:

>> logspace(d1,d2,n)

where n is the number of points to be generated.

For example, to generate 100 points between 1 rad/sec and 1000 rad/sec, use:

>> W=logspace(0,3,100)

To draw the Bode plot, we use the command >> sys=tf(num,den)

>> bode(sys,W)

To display the gain and phase margins >> margin(sys)

Suppose we need to draw the Bode plot for the control system:

𝐺𝐻(𝑠) =25(𝑆 + 5)

𝑆(𝑆2 + 3𝑆 + 10)(𝑆 + 50)

So we write the following Matlab code

The magnitude and phase plots are as shown below in Fig. 10, and it is clear that this

is an unstable system.




Fig. 10, Magnitude and phase plots

Example (1):

Obtain the phase and gain margins of the system shown below, for the two cases K=10

and K=100.

First, we draw the Bode plot at K=10 as shown in Fig. 11, and calculate the Gain and

Phase margins as:

G.M. = 8 dB (+ve) and Phase Margin = 21 (+ve). This gives Stable System




Fig. 11, Bode plot at K=10

Now, the system gain is changed to 100.

Changing the system gain change the magnitude plot up or down depending on K. But

the phase plot is not affected at all.

Therefore, no need to re-plot the gain plots at the new gain, but we can modify each

point on the magnitude plot up by 20 dB (20 log100 - 20 log10) as shown in Fig. 12.




Fig. 12, Bode plot at K=100

From Fig. 11, the Gain and Phase margins are:

G.M. = 12 dB (-ve) and Phase Margin = 30 (-ve). This gives Unstable System

6. Finding T.F. & Steady-State Error Coefficient from Bode Plot

The asymptotic Bode plot must have slopes of multiples of 20dB/decade. If the

slope is changed from -20 to -40 dB/decade at =1, this means there is one pole at 1

Also, If the slope is changed from -20 to -60 dB/decade at =2, this means there is

quadratic poles at 2




On the other hand, if the slope is changed from -20 to 0 dB/decade at =1, this

means there is one zero at 1. Also, If the slope is changed from -20 to +20 dB/decade

at =2, this means there is quadratic zeros at 2

The type of the system determines the slope of the log-magnitude curve at low

frequencies. Now, to calculate how many poles at origin (system type) and the system

gain (K), we need to examine the first portion of the plot as follows:

6.1 Type zero systems:

It follows that the low-frequency asymptote is a horizontal line at 20 log Kp dB.

6.2 Type one systems

The intersection of the initial –20-dB/decade segment (or its extension) with the line ω

= 1 has the magnitude 20 log Kv

The intersection of the initial –20-dB/decade segment (or its extension) with the 0-dB

line has a frequency numerically equal to Kv.




6.3 Type two systems

The intersection of the initial –40-dB/decade segment (or its extension) with the ω = 1

line has the magnitude of 20 log Ka.

The frequency ωa at the intersection of the initial –40-dB/decade segment (or its

extension) with the 0-dB line gives the square root of Ka numerically.

Example (2)

Consider the magnitude plot of the open-loop T.F. given below:

Obtain:

a) The steady-state error coefficient

b) The open-loop T.F.

Referring to the first section in the magnitude plot given above, we find the initial

slope is −40 dB/dec, this means this system is type 2.

We can calculate the acceleration error coefficient Ka from the point of intersection

between the extension of −40 dB/dec line with 0 dB line which is 3.0 rad/s




√𝐾𝑎 = 3 → 𝐾𝑎 = 9

Also, we can get Ka by obtaining the magnitude at ω = 1

20 log(𝐾𝑎) = 19.1 → 𝐾𝑎 = 9

Therefore, the system gain K = Ka = 9 and the open-loop T.F. is given as:

𝐺𝐻(𝑗𝜔) =9 (1 +

𝑗𝜔1

)

𝑗𝜔2 (1 +𝑗𝜔4

)

Example (3)

Consider the magnitude plot of the open-loop T.F. given below:

Obtain:

a) The steady-state error coefficient

b) The open-loop T.F.

Referring to the first section in the magnitude plot given above, we find the initial

slope is 0 dB/dec, this means this system is type 0.

We can calculate the position error coefficient Kp as

20 log 𝐾𝑝 = 25 → 𝐾𝑝 = 17.7828

Therefore, the system gain K = Kp = 17.7828 and the open-loop T.F. is given as:

𝐺𝐻(𝑗𝜔) =17.7828 (1 +

𝑗𝜔5

) (1 +𝑗𝜔8

)

(1 +𝑗𝜔4

) (1 +𝑗𝜔7

) (1 +𝑗𝜔9

)

7. Design Via Gain Adjustment

Referring to Fig. below, we see that if we desire certain phase margin, represented by

CD, we would have to raise the magnitude curve by AB. Thus, a simple gain

adjustment can be used to design phase margin.




On the other hand, if we desire certain gain margin, represented by AC, and from the

system Bode plot we find the actual gain margin is AB (as shown in Fig. below).

Therefore, we would have to decrease the magnitude curve by BC. Thus, a simple gain

adjustment can be used to design gain margin also.

Design Procedure:

• Assume the system gain K = 1 (At this value, the magnitude plot is not affected

as 20 log(1) = 0. And the phase plot is not affected as the 1 = 0˚)

• Draw the system Bode magnitude and phase plots.

• Based on the required Gain Margin or required Phase Margin, determine the

amount of increment or decrement in the magnitude plot.




Example (4)

For the open loop transfer function given below:

GH(s) =500K (S + 2)

S (S + 1)(S2 + 5S + 100) (S = jω)

a) Construct the Bode plots,

b) The Gain crossover frequency and the Gain Margin,

c) The phase crossover frequency and the Phase Margin,

d) Design the value of K to change the gain crossover frequency to 1 rad/s, then find

the corresponding Gain and Phase Margins,

e) Design the value of K to give a phase margin of 54, then find the gain crossover

frequency and the corresponding Gain Margin,

f) Design the value of K to give a critical stable system.

For the transfer function given below:

𝐻(𝑠) =500 𝐾(𝑆 + 2)

𝑆 (𝑆 + 1)(𝑆2 + 5𝑆 + 100)

we put the T.F. in Bode form as:

𝐻(𝑠) =10𝐾 (1 +

𝑗𝜔2

)

𝑗𝜔 (1 +𝑗𝜔1

) {1 +5𝑗𝜔100

+ (𝑗𝜔10

)2

}

Assuming K =1,

The Bode gain = 20 log (10) = 20 dB

The total magnitude and total phase are plotted as given in semi-log paper.

a) Bode plots are sketched as given in semilog paper.

From Bode plots:

b) The gain crossover frequency = 4.8 rad/s & Gain Margin = 2dB

c) The phase crossover frequency = 6.4 rad/s & Phase Margin = 14

d) To change the gain crossover frequency to 1.0 rad/s, the total magnitude at this

point must be zero dB. But from Bode plot, it is found that the total magnitude is +20

dB, so it is recommended to add -20 dB at that point




20 log 𝐾 = −20

K = 0.1

At this value of K, Gain Margin= 22 dB and Phase Margin = 72

e) To change the phase margin to 54, this mean the gain crossover frequency must be

changed to 1.8 rad/s, and the total magnitude at this point must be zero dB. But from

Bode plot, it is found that the total magnitude is +10 dB, so it is recommended to add

-10 dB at that point

20 log 𝐾 = −10

K = 0.316, At this value of K, Gain Margin= 12 dB




Example (5)

A system for controlling the performance of an aircraft in the lateral mode is

equivalent to a single loop with open loop transfer function:

𝐺(𝑆)𝐻(𝑆) = 250𝐾(𝑆 + 20)

𝑆(𝑆 + 5)(𝑆 + 10)2

a) Draw Bode plots at gain K = 1 and discuss the system stability.

b) Find the value of K to get phase margin of 45

c) Find the value of K to get gain margin of 20 dB

𝐺(𝑆)𝐻(𝑆) = 250𝐾(𝑆 + 20)

𝑆(𝑆 + 5)(𝑆 + 10)2

𝐺(𝑗𝜔)𝐻(𝑗𝜔) = 250 × 20𝐾(1 +

𝑗𝜔20

)

5 × 100 𝑗𝜔 (1 +𝑗𝜔5

) (1 +𝑗𝜔10

)2

𝐺(𝑗𝜔)𝐻(𝑗𝜔) = 10𝐾(1 +

𝑗𝜔20

)

𝑗𝜔 (1 +𝑗𝜔5

) (1 +𝑗𝜔10

)2

At K = 1,

For the Bode gain 10, it can be represented as 20 log (10) = 20 dB

Pole at origin: the magnitude is represented by a straight line with slope of -20

dB/decade and intersect the 0dB line at ω=1 and extended to intersect the vertical axis.

But the phase is represented by straight line parallel to horizontal axis with constant

value at -90.

Real zero (S+20): the magnitude is represented by a straight line with slope of +20

dB/decade and intersect the 0dB line at ω=20. The phase is represented as:

𝜑 = {0 𝑎𝑡 𝜔 < 2

+45 𝑎𝑡 𝜔 = 20+90 𝑎𝑡 𝜔 > 200

Real Pole (S+5): the magnitude is represented by a straight line with slope of -20

dB/decade and intersect the 0dB line at ω=5. The phase is represented as:

𝜑 = {0 𝑎𝑡 𝜔 < 0.5−45 𝑎𝑡 𝜔 = 5−90 𝑎𝑡 𝜔 > 50

Repeated Real Pole (S+10)2: the magnitude is represented by a straight line with

slope of -40 dB/decade and intersect the 0dB line at ω=10. The phase is represented

as:




𝜑 = {0 𝑎𝑡 𝜔 < 1

−90 𝑎𝑡 𝜔 = 10−180 𝑎𝑡 𝜔 > 100

From Bode plot shown on semi-log paper, we find that

a) ω1 = 6.6 rad/s → Phase margin = -13

ωc = 4.7 rad/s → Gain margin = -5 dB, this mean the system is unstable ##

b) At Phase margin = 45, the system magnitude is +20 dB

Then 20 log (K) = -20 → K = 0.1 ##

c) To get Gain margin of 20 dB,

20 log (K) = -(20+5) K = 0.05623 ##

7. Closed-Loop Frequency Response

Consider the second-order system given in Fig. 13, whose open-loop T.F. is given as:

G(s)H(s) =𝜔𝑛

2

𝑆(𝑆 + 2𝜁𝜔𝑛)

Also, the closed-loop T.F. is given as:

𝐶(𝑠)

𝑅(𝑠)=

𝜔𝑛2

𝑆2 + 2𝜁𝜔𝑛𝑆 + 𝜔𝑛2

Fig. 13, Second-order system

To obtain the closed-loop frequency response, replace each S in the closed-loop T.F

by jω as follows:

𝐶(𝑗𝜔)

𝑅(𝑗𝜔)=

𝜔𝑛2

(𝑗𝜔)2 + 2𝜁𝜔𝑛(𝑗𝜔) + 𝜔𝑛2

=𝜔𝑛

2

𝜔𝑛2 − 𝜔2 + 𝑗2𝜁𝜔𝑛𝜔

The magnitude [M(jω)] of the above closed-loop T.F. is:

|𝑀(𝑗𝜔)| =𝜔𝑛

2

√(𝜔𝑛2 − 𝜔2)2 + 4𝜁2𝜔𝑛

2𝜔2

The magnitude plot obtained from the representation of M(jω) with respect to ω is

shown in Fig. 14.




Fig. 14, Magnitude plot of closed-loop T.F.

From the above figure, there is a peak value which is 20×log(Mp) in dB or Mp in

actual scale. To obtain this peak value (Mp), we get the derivative dM(jω)/dω and

equating it to zero as follows:

𝑑𝑀(𝑗𝜔)

𝑑𝜔=

−0.5𝜔𝑛2(4𝜔3 − 4𝜔𝑛

2𝜔 + 8𝜁2𝜔𝑛2𝜔)

(𝜔𝑛2 − 𝜔2)2 + 4𝜁2𝜔𝑛

2𝜔2= 0

The frequency at which peak value is occurred is called peak frequency (ωp)

4𝜔𝑝3 − 4𝜔𝑛

2𝜔𝑝 + 8𝜁2𝜔𝑛2𝜔𝑝 = 0

𝜔𝑝(𝜔𝑝2 + 2𝜁2𝜔𝑛

2 − 𝜔𝑛2) = 0

𝜔𝑝2 = 𝜔𝑛

2(1 − 2𝜁2) → 𝜔𝑝 = 𝜔𝑛√1 − 2𝜁2

To get the peak magnitude Mp, substitute with the value of ωp in the equation

𝑀𝑝 = |𝑀(𝑗𝜔)|𝜔=𝜔𝑝=

𝜔𝑛2

√(𝜔𝑛2 − 𝜔𝑛

2(1 − 2𝜁2))2

+ 4𝜁2𝜔𝑛2𝜔𝑛

2(1 − 2𝜁2)

𝑀𝑝 =1

2𝜁√1 − 𝜁2

The effect of changing damping ratio on Mp and ωp is shown in Fig. 15.

Fig. 15, Peak amplitude at different damping ratios




The Band Width frequency (ωBW), is frequency at which the magnitude response curve

is 3 dB down (on log scale) (or 1/√2=0.7071 on normal scale) from its value at zero

frequency.

1

√2× |𝑀(𝑗𝜔)|𝜔=0 =

1

√2× 1 = 0.7071 =

𝜔𝑛2

√(𝜔𝑛2 − 𝜔𝐵𝑊

2)2 + 4𝜁2𝜔𝑛2𝜔𝐵𝑊

2

𝜔𝐵𝑊 = 𝜔𝑛√1 − 2𝜁2 + √4𝜁4 − 4𝜁2 + 2

Example (6)

Find the closed-loop bandwidth required for 20% overshoot and 2-seconds settling

time.

The maximum overshoot is 0.2, and as we know it is given by:

0.2 = 𝑒−

𝜋 𝜁

√1− 𝜁2 → 𝜁 = 0.45595

The settling time (based on ±2% tolerance) is given as:

𝑇𝑠 = 2 = 4

𝜁 𝜔𝑛 → 𝜔𝑛 = 4.38645 𝑟𝑎𝑑/𝑠

The BW frequency is calculated from:

𝜔𝐵𝑊 = 𝜔𝑛√1 − 2𝜁2 + √4𝜁4 − 4𝜁2 + 2

𝜔𝐵𝑊 = 4.38645√1 − 2 × 0.20789 + √4 × 0.04322 − 4 × 0.20789 + 2 = 5.79 𝑟𝑎𝑑/𝑠

Example (7)

For the closed-loop T.F given below,

𝐶(𝑠)

𝑅(𝑠)=

5

𝑆2 + 2𝑆 + 5

Calculate the peak amplitude (Mp), frequency (ωp) at which Mp is occurred and the

band width (BW) of the closed-loop frequency response. Then give a free-hand sketch

for the closed-loop frequency response.

Since the system is 2nd order system, we can use the equations given above. But, first

we need to calculate ζ and ωn. By comparing the characteristic eqn. of the given

system with the standard one:




𝑆2 + 2ζ𝜔𝑛𝑆 + 𝜔𝑛2 = 0

We obtain that:

𝜔𝑛2 = 5 → 𝜔𝑛 = √5

2ζ𝜔𝑛 = 2 → ζ = 1/√5

𝜔𝑝 = 𝜔𝑛√1 − 2𝜁2 = √5√1 − 2 × 0.2 = √3 = 1.7321

𝑀𝑝 =1

2𝜁√1 − 𝜁2=

1

2

√5√1 − 0.2

=1

0.8= 1.25

𝜔𝐵𝑊 = 𝜔𝑛√1 − 2𝜁2 + √4𝜁4 − 4𝜁2 + 2 = √5√0.8 + √1.36 = 3.1354

Example (8)

The open-loop T.F of a unity-feedback control system is

𝐺(𝑠) =𝐾

𝑆(𝑆 + 𝑎)

And the closed-loop frequency response |𝑀(𝑗𝜔)| versus ω is given in figure below.

Find the value of K and a, then calculate the band width (BW)

The closed-loop T.F. is:

𝐶(𝑠)

𝑅(𝑠)=

𝐾

𝑆2 + 𝑎𝑆 + 𝐾

By comparing the characteristic eqn. of the given system with the standard one:

𝑆2 + 2ζ𝜔𝑛𝑆 + 𝜔𝑛2 = 0




We obtain that:

𝜔𝑛2 = 𝐾

2ζ𝜔𝑛 = 𝑎

From the closed-loop frequency response, Mp = 1.5 and ωp = 2 rad/s

𝑀𝑝 = 1.5 =1

2𝜁√1 − 𝜁2

3𝜁√1 − 𝜁2 = 1

9𝜁2(1 − 𝜁2) = 1

9𝜁4 − 9𝜁2 + 1 = 0

Let ζ2 = x, therefore, ζ = √x

9𝑥2 − 9𝑥 + 1 = 0

x = 0.87278, → ζ = 0.93423

x = 0.127322, → ζ = 0.3568

𝜔𝑝 = 2 = 𝜔𝑛√1 − 2𝜁2

At ζ = 0.93423,

𝜔𝑛 =2

√1 − 2𝜁2=

2

√1 − 2 × 0.87278 𝑡ℎ𝑖𝑠 𝑔𝑖𝑣𝑒𝑠 𝑖𝑚𝑎𝑔𝑖𝑛𝑎𝑟𝑦 𝑣𝑎𝑙𝑢𝑒

Therefore, ζ = 0.93423 is rejected.

𝜔𝑛 =2

√1 − 2𝜁2=

2

√1 − 2 × 0.127322= 2.3166 𝑟𝑎𝑑/𝑠

ζ = 0.3568 and ωn = 2.3166

𝐾 = 𝜔𝑛2 = 5.3665

𝑎 = 2ζ𝜔𝑛 = 2 × 0.3568 × 2.3166 = 1.6531

𝜔𝐵𝑊 = 𝜔𝑛√1 − 2𝜁2 + √4𝜁4 − 4𝜁2 + 2 = 2.3166√0.745356 + √1.5556 = 3.27 𝑟𝑎𝑑/𝑠

Example (9)

The open-loop T.F of a unity-feedback control system is,

𝐺(𝑠) =60

𝑆(𝑆 + 2)(𝑆 + 6)

Calculate the peak amplitude (Mp), frequency at which Mp is occurred (ωp) and the

band width (BW) of the closed-loop frequency response.

The closed-loop T.F. is: 𝐶(𝑠)

𝑅(𝑠)=

60

𝑆(𝑆 + 2)(𝑆 + 6) + 60=

60

𝑆3 + 8𝑆2 + 12𝑆 + 60




This is a 3rd order system, so we can’t use the equations given as a function of ζ and ωn 𝐶(𝑗𝜔)

𝑅(𝑗𝜔)=

60

(𝑗𝜔)3 + 8(𝑗𝜔)2 + 12(𝑗𝜔) + 60=

60

(60 − 8𝜔2) + 𝑗(12𝜔 − 𝜔3)

𝑀(𝑗𝜔) = |𝐶(𝑗𝜔)

𝑅(𝑗𝜔)| =

60

√(60 − 8𝜔2)2 + (12𝜔 − 𝜔3)2

𝑑𝑀(𝑗𝜔)

𝑑𝜔=

−60 × 0.5(… )−0.5{256𝜔3 − 1920𝜔 + 288𝜔 + 6𝜔5 − 96𝜔3}

(60 − 8𝜔2)2 + (12𝜔 − 𝜔3)2= 0

256𝜔3 − 1920𝜔 + 288𝜔 + 6𝜔5 − 96𝜔3 = 0

6𝜔5 + 160𝜔3 − 1632𝜔 = 0

𝜔4 + 26.6667𝜔2 − 272 = 0

Let ω2 = x, → ω = √x

𝑥2 + 26.6667𝑥2 − 272 = 0

x = -34.5413 (rejected)

x = 7.874632 (accepted) → ωp = 2.8062

|𝐶(𝑗𝜔)

𝑅(𝑗𝜔)|

𝜔𝑝

=60

√(60 − 8 × 7.874632)2 + 7.874632(12 − 7.874632)2= 5.0175

8. Relation Between Open-& Closed-Loop Frequency Responses (Nichole Chart)

The Bode plot is generally constructed for an open loop transfer function of a system.

In order to draw the Bode plot for a closed loop system, the transfer function has to be

developed, and then factorized to its poles and zeros. This process is tedious and

cannot be carried out without the aids of a powerful calculator or a computer.

Nichols, developed a simple process by which the unity feedback closed-loop

frequency response of a system can be easily deduced from the open-loop transfer

function. This approach is outlined as follows:

Consider a unity feedback control system given in Fig. 16.

Fig. 16, Unity feedback control system




The closed-loop T.F. is:

𝐶(𝑠)

𝑅(𝑠)=

𝐺(𝑠)

1 + 𝐺(𝑠)

The polar plot of G(jω) in the GH plane is shown in Fig. 17. Where the vector OA

represents the magnitude and angle of G(jω) in G-plane.

Fig. 17, Polar plot for G(jω) represented in G-plane

Assuming that the real part of G(jω) is x(ω) and the imaginary part of G(jω) is y(ω).

Therefore, the closed-loop T.F. is expressed as:

𝐶(𝜔)

𝑅(𝜔)=

𝑥(𝜔) + 𝑗𝑦(𝜔)

1 + 𝑥(𝜔) + 𝑗𝑦(𝜔)= 𝑀∠∅

𝑀 =√𝑥2 + 𝑦2

√(1 + 𝑥)2 + 𝑦2

𝑀2 =𝑥2 + 𝑦2

(1 + 𝑥)2 + 𝑦2

𝑀2(1 + 𝑥)2 + 𝑀2𝑦2 = 𝑥2 + 𝑦2

Expanding and collecting similar terms:

(1 − 𝑀2) 𝑥2 − 2𝑀2𝑥 − 𝑀2 + (1 − 𝑀2)𝑦2 = 0

At M = 1; x (real part of G) = −0.5

At M ≠ 1; the above equation is divided by (1−M2) and is rewritten as:

𝑥2 +2𝑀2

𝑀2 − 1𝑥 +

𝑀2

𝑀2 − 1+ 𝑦2 = 0




Completing square;

(𝑥 +𝑀2

𝑀2 − 1)

2

+ 𝑦2 =𝑀2

𝑀2 − 1

This is an equation of a circle with radius

𝑀

𝑀2 − 1

And its center at

𝑥 = −𝑀2

𝑀2 − 1, 𝑦 = 0

The representation of the circle equation given above at different values of M is given

in Fig. 18. Thus, if the polar frequency response of an open-loop function, G(s), is

plotted and superimposed on top of the constant M circles, the closed-loop magnitude

frequency response is determined by each intersection of this polar plot with the

constant M circles.

Fig. 18, Constant M circles

The phase angle of the closed-loop T.F is given by:

∅ = 𝑡𝑎𝑛−1𝑦

𝑥− 𝑡𝑎𝑛−1

𝑦

𝑥 + 1




Taking tan for both sides:

tan(∅) = 𝑁 =

𝑦𝑥

−𝑦

𝑥 + 1

1 +𝑦𝑥

.𝑦

𝑥 + 1

=𝑦(𝑥 + 1) − 𝑦(𝑥)

𝑥(𝑥 + 1) + 𝑦2=

𝑦

𝑥2 + 𝑥 + 𝑦2

𝑥2 + 𝑥 + 𝑦2 −1

𝑁𝑦 = 0

Completing squares:

(𝑥 +1

2)

2

+ (𝑦 −1

2𝑁)

2

= (1

2)

2

+ (1

2𝑁)

2

This is an equation of a circle with radius

√(1

2)

2

+ (1

2𝑁)

2

And its center at

𝑥 =1

2, 𝑦 =

1

2𝑁

The representation of the circle equation given above at different values of N is given

in Fig. 19. Superimposing a unity feedback, open-loop frequency response over the

constant N circles yields the closed-loop phase response of the system.

Both M and N circles are represented in one figure to get the magnitude and angle

easily as shown in Fig. 20. The main disadvantage of using the M and N circles is that

changes of gain in the open-loop transfer function, G(s), cannot be handled easily. For

example, in the Bode plot, a gain change is handled by moving the Bode magnitude

curve up or down an amount equal to the gain change in dB scale. Since the M and N

circles are not dB plots, changes in gain require each point of G(jω) to be multiplied in

length by the increase or decrease in gain.

Another presentation of the M and N circles, called a Nichols chart, displays the

constant M circles in dB, so that changes in gain are as simple to handle as in the Bode

plot. A Nichols chart is shown in Fig. 21.




Fig. 19, Constant N circles

Fig. 20, Constant M and N circles




Fig. 21, Nichole Chart




Sheet 10 (Bode Plots)

(1) For the following transfer functions, sketch the Bode plots:

)10)(1(

1)(

++=

sssG

)20)(1(

)10()(

++

+=

ss

ssG

)502(

1)(

2 ++=

sssG

)20)(1(

)10(10)(

++

+=

sss

ssG

)5012)(1(

)5()(

2 +++

−=

sss

ssG

)10)(1(

)5(1)(

2 ++

+=

sss

ssG

(2) For each of the following transfer functions, sketch the Bode diagram and

determine the gain crossover frequency (that is, the frequency at which the

magnitude of G(j)=20 log G(j)= 0 dB, and the phase crossover frequency

(that is, the frequency at which the phase angle of G(j)=-180:

)2)(10(

1000)(

++=

sssG

)20)(1.0(

100)(

2 +++=

ssssG

)100)(1(

)10(50)(

++

+=

ss

ssG

)500)(80)(5(

)5014(1000)(

2

+++

++=

sss

sssG

(3) Consider the non-unity feedback system in Figure (1), where the controller

gain is K=2. Sketch the Bode plot of the open loop transfer function. Determine

the phase angle of the open loop transfer function when the magnitude equals

to 0 dB.

Figure (1) Non-unity feedback system with controller gain K

(4) Consider the system given in Figure (2) where




• )10(

)5()(

+

+=

s

sKsGc

• )22(

1)(

2 ++=

ssssGp

• 1)( =sH

(a) Find K such that the velocity error coefficient Kv=10 (b) Draw the Bode plot of the open-loop system (c) From the Bode plot, find the frequency corresponding to 0 dB gain

(d) From the Bode plot, find the frequency corresponding to -180 phase

Figure (2) A closed-loop control system

(5) The asymptotic log-magnitude curves for two transfer functions are given in

Figure (3). Sketch the corresponding asymptotic phase angle curves for each system.

Determine the transfer function for each system.

Figure (3) Log-magnitude curves

(6) A position control system may be constructed by using an AC motor and AC

components, as shown in Figure (4-a). To measure the open-loop frequency

response, we simply disconnect X from Y and X' from Y' and then apply a sinusoidal

modulation signal generator to the Y - Y' terminals and measure the response at X -

X'. The resulting frequency response of the loop transfer function L(j)=

Gc(j)G(j)H(j), is shown in Figure (4-b). Determine the transfer function L(j).




Figure (4) (a) AC motor control, (b) Frequency response

(7) A helicopter with a load on the end of a cable is shown in Figure (5-a). The

position control system is shown in Figure (5-b), where the visual feedback is

represented by H(s). Sketch the Bode diagram of G(j)H(j).

Figure (5) A helicopter feedback control system




(8) Sketch Bode plot of a system with transfer function

)22)(21(

)1.01()(

2 +++

+=

ssss

sKsG

(a) K=10 (b) K=20 (c) Compare plots obtained in (a) and (b)

(9) A system has the loop transfer function

)49/7/1)(21(

)5/1(5.2)(

2ssss

ssG

+++

+=

Plot the Bode diagram. Show that the phase margin is approximately 28 and that the

gain margin is approximately 21 dB.

(10) Consider a unity feedback system with the loop transfer function

)04.024.01)(21(

)4.01(10)()(

2ssss

ssHsG

+++

+=

(a) Plot the Bode diagram,

(b) Find the gain margin and phase margin.

(11) The experimental Oblique Wing Aircraft (OWA) has a wing that pivots, as shown

in Figure (6). The wing is in the normal unskewed position for low speeds and can

move to a skewed position for improved supersonic flight. The aircraft control system

loop transfer function is

])8/()20/(1)[21(

)5.01()()(

2ssss

sKsGsG c

+++

+=

Figure (6) The Oblique Wing Aircraft, top and side




(a) Sketch the Bode diagram when K=4,

(b) Find the gain margin and phase margin,

(c) Find the value of K for critical stable system.

(12) Consider a unity feedback system with the following open-loop transfer function:

)2)(1()(

++=

sss

KsG

(a) For K=4, show that the gain margin is 3.5 dB, (b) If we wish to achieve a gain margin equal to 16 dB, determine the value of the

gain K.

(13) A closed-loop system, as shown in Figure (7) has H(s)=1 and

)02.01)(02.01()()(

sss

KsGsG c

++=

Figure (7) Feedback control system

(a) Select a gain K so that the steady-state error for a ramp input is 10% of the magnitude of the ramp function A, where r(t) = At, t> 0,

(b) Plot the Bode plot of Gc(s)G(s), and determine the phase and gain margins.

(14) An electric carrier that automatically follows a tape track laid out on a factory

floor. Closed-loop feedback systems are used to control the guidance and speed of

the vehicle. The block diagram of the steering system is shown in Figure (8). Select a

gain K so that the phase margin is approximately 30° .

Figure (8) Feedback steering control system




(15) Consider the system shown in Figure (9).

Figure (9) Feedback control system

(a) Draw a Bode diagram of the open-loop transfer function,

(b) Determine the value of the gain K such that the phase margin is 50 , (c) What is the gain margin of this system with this gain K?

References:

[1] Bosch, R. GmbH. Automotive Electrics and Automotive Electronics, 5th ed. John Wiley & Sons Ltd., UK,

2007.

[2] Franklin, G. F., Powell, J. D., and Emami-Naeini, A. Feedback Control of Dynamic Systems. Addison-

Wesley, Reading, MA, 1986.

[3] Dorf, R. C. Modern Control Systems, 5th ed. Addison-Wesley, Reading, MA, 1989.

[4] Nise, N. S. Control System Engineering, 6th ed. John Wiley & Sons Ltd., UK, 2011.

[5] Ogata, K. Modern Control Engineering, 5th ed ed. Prentice Hall, Upper Saddle River, NJ, 2010.

[6] Kuo, B. C. Automatic Control Systems, 5th ed. Prentice Hall, Upper Saddle River, NJ, 1987.

chapter 12 frequency response analysis (bode ... shoubra...• plot asymptotic approximations to the...

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